用C语言编程实现生产者-消费者问题，用word文档提交可运行的源代码，并给出运行效果截图。

生产者-消费者问题是经典的并发控制问题，通常用于演示线程同步的概念。在C语言中，我们可以使用互斥锁（mutex）和条件变量（condition variables）来解决这个问题。下面是一个简化版的生产者-消费者模型的示例，我将提供伪代码解释，但由于此处无法创建实际的word文档、源代码文件或截图，你可以在本地尝试编写并测试。

#include <pthread.h>
#include <semaphore.h>

#define BUFFER_SIZE 5
#define SEMAPHORE_PRODUCER 1
#define SEMAPHORE_CONSUMER 2

typedef struct {
    int buffer[BUFFER_SIZE];
    int count;
    int in;
    int out;
    pthread_mutex_t mutex;
    sem_t producer_semaphore;
    sem_t consumer_semaphore;
} Buffer;

void* producer(void* arg) {
    Buffer* buffer = (Buffer*)arg;
    while (1) {
        // 生产者获取资源
        pthread_mutex_lock(&buffer->mutex);
        if (buffer->count == BUFFER_SIZE) {
            sem_wait(&buffer->producer_semaphore); // 阻塞直到有空位
        }
        buffer->buffer[buffer->in] = rand(); // 生产元素
        buffer->count++;
        buffer->in = (buffer->in + 1) % BUFFER_SIZE; // 循环索引
        pthread_mutex_unlock(&buffer->mutex);

        // 提醒消费者产品可用
        sem_post(&buffer->consumer_semaphore);
    }
}

void* consumer(void* arg) {
    Buffer* buffer = (Buffer*)arg;
    while (1) {
        // 消费者等待产品
        pthread_mutex_lock(&buffer->mutex);
        if (buffer->count == 0) {
            sem_wait(&buffer->consumer_semaphore); // 阻塞直到有产品
        }
        printf("Consumed %d\n", buffer->buffer[buffer->out]); // 消费并打印
        buffer->count--;
        buffer->out = (buffer->out + 1) % BUFFER_SIZE; // 循环索引
        pthread_mutex_unlock(&buffer->mutex);

        // 通知生产者空间已满
        sem_post(&buffer->producer_semaphore);
    }
}

int main() {
    Buffer buffer;
    init_buffer(buffer);
    pthread_t producer_thread, consumer_thread;
    pthread_create(&producer_thread, NULL, producer, &buffer);
    pthread_create(&consumer_thread, NULL, consumer, &buffer);
    pthread_join(producer_thread, NULL);
    pthread_join(consumer_thread, NULL);
    return 0;
}